C3 Trigonometry

Original post by Mr Maths

by this I assume you mean get all the values in the new boundaries (original boundaries multiplied by 3/2) and then divide each value by 3/2 afterwards

don't want to confuse to OP :smile:

Original post by Fazeel 123

Then it would be tan(3x/2)=-1
tan-1(-1)=3x/2
times by 2 and divide by 3 to get X and get the rest of the values in the range given in the question

Sorry everyone I was unclear. It is that equation equal to zero, and then find all the angles between 0 and 360 that satisfy the equation.

Original post by hassi94

That's not a question, what's the actual question? :smile:

Original post by Fazeel 123

Is it 1+tan(3x/2)=0

Original post by steve2005

Is there supposed to be an equation somewhere?

Original post by GreenLantern1

Can you please help me on this? This one I am really confused about:

1+ \tan(3x/2)

Please explain in detail.I don't think we have even covered it in class, yet we have been set to do it. Could you explain why you do some things.

Reply 7

1+ tan3x/2 =0 0<x<360
tan3x/2=-1
tan3x/2 = -1
3x/2 = tan^-1(-2) 0<3x/2<540
Then add 180 to your first value and then keep adding 180 until you get the last value that is <540
Now divide all the values by 3 and times by 2 and check if any of them don't fit in the original range and if so then don't count them.

(edited 12 years ago)

Reply 8

Original post by GreenLantern1

Sorry everyone I was unclear. It is that equation equal to zero, and then find all the angles between 0 and 360 that satisfy the equation.

Surely you've done this in C2?

Inverse tan, and since 0<x<360 then 0<3x/2<540 so find all the solutions in that range, then divide by 3/2 to get x =..

Reply 9

Original post by TLK

1+ tan3x/2 =0 0<x<360
tan3x/2=-1
tan3x = -2
3x = tan^-1(-2) 0<3x<1080
Then add 180 to your first value and then keep adding 180 until you get the last value that is <1080
Now divide all the values by 3 and check if any of them don't fit in the original range and if so then don't count them.

This is incorrect.

Reply 10

Original post by hassi94

This is incorrect.

Ah, sorry I didn't look at the 2 and though it was outside the bracket :colondollar:

.
Thanks, I've changed it now

Reply 11

So is it 210 and 30 degrees?

Reply 12

2 cot y = 3 cos y

How do I do this!!

Reply 13

Original post by GreenLantern1

So is it 210 and 30 degrees?

How did you get that? FYI that's incorrect.

Try do this question before you move onto any more.

Reply 14

Original post by GreenLantern1

2 cot y = 3 cos y

How do I do this!!

2 cot y = 3 cos y
2 1/tany = 3 cos y
2 cosy/siny = 3 cos y
2/siny = 3
2/3 = siny

Hopefully I'm right first time this time :colondollar:

Reply 15

Original post by TLK

2 cot y = 3 cos y
2 1/tany = 3 cos y
2 cosy/siny = 3 cos y
2/siny = 3
2/3 = siny

Hopefully I'm right first time this time :colondollar:

I'm afraid not. :tongue:

By doing what you did you've lost the solution cos y = 0

I'll help you with this OP once you've got the first one. :smile:

Reply 16

Original post by hassi94

How did you get that? FYI that's incorrect.

Try do this question before you move onto any more.

I don't understand what I've done wrong then :|. Are either of the values correct?

Reply 17

Original post by hassi94

I'm afraid not. :tongue:

By doing what you did you've lost the solution cos y = 0

I'll help you with this OP once you've got the first one. :smile:

, I get 90 when inverse cos y = 0 but when I try to sub it back in for 2/tan(90) I get maths error??

Reply 18